If f(x) = ^2 x + ^2 x, then - Vidyadip

MCQ

If $f(x) = {\cos ^2}x + {\sec ^2}x,$ then

A
$f(x) < 1$
B
$f(x) = 1$
C
$1 < f(x) < 2$
✓
$f(x) \ge 2$

✓

Answer

Correct option: D.

$f(x) \ge 2$

d
(d) Since ${\left( {x - \frac{1}{x}} \right)^2} \ge 0,\,\,{\rm{\rlap{--} V}}\,\,x \in R,$

we have ${x^2} + \frac{1}{{{x^2}}} \ge 2$ and

Hence, $f(x) = {\cos ^2}x + \frac{1}{{{{\cos }^2}x}} \ge 2$.

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